Catsidhe

1. Set a problem for homework.
   
2. Provide a suggested method for finding a solution of, essentially, "randomly shuffle these numbers until it kinda looks right".
   
3. ...
   
4. Profit! End up with children who are frustrated and angered by the sight of numbers, and have little to no idea that there are ways in which this sort of problem can be approached, let alone relatively simple and rigorous ways to prove them correct, let alone that the solution raises all sorts of other questions, which can themselves be answered...

The actual problem was "Take the nine numbers 2 to 10, and arrange them in three groups of three so that each group adds to the same number."

The suggested approach was to "write the numbers on pieces of paper, and arrange them into the right groups."

No, seriously, the suggested approach was to randomly shuffle the numbers until they (magically) come out in the right order. Personally, I'm wondering if there is a worse possible approach to the problem.

When I sat down with Miss A to approach this, my first question was: so, what is the number they have to add up to?

What you're looking for is

x = a + b + c
  = d + e + f
  = g + h + i

So the first thing to notice is that

a + b + c + d + e + f + g + h + i = 3x

The sum of 2..10 is 54, so the answer to each group of three must be 54/3 = 18.

So then we need an algorithm to fill in the blanks. Start with the biggest number, so

18 = 10 + b + c

b≠9, because that is already too big. And while 10+8 = 18, that only works if c=0, which isn't an available value. Neither is 1, so b≠7. So by elimination, we have a=10, b=6, c=2. Then do the same with the remaining numbers (d=9, e=5, f=4), and the remaining three must be g, h and i. Luckily, when you check, they are.

There was a secondary part to do the same thing with the set B = [3..11]. And yes, we showed that the algorithm still works. Only now the sum to each group is 21.

Hang on, 21 = 18+3, and we're dealing with groups of three... that can't be a coincidence, can it? It turns out, if you compare the ordered sets, then you see that each number B_x is just A_x+1. So if each number has 1 added, then each group must have 3 added to the total for it to work out.

And if we've just solved this problem for the set N₂ = [2..10], and for N₂₊₁, then we've demonstrated that the solution will work for N_x, where x is any positive integer. So for the set [1..9], the sum to each group should be 15... and when you check, it is.

But wait... what we've got can be drawn in a grid

10	6	2	=18
9	5	4	=18
8	7	3	=18

If we re-arrange the numbers within each row, then we get

10	6	2	=18
5	4	9	=18
3	8	7	=18
= 18	= 18	= 18

And if you do a bit of matrix manipulation, then you get a Magic Square, where the rows, columns and diagonals all add up to the same magic number.

And we've proved that this pattern is a Magic Square whether you pick your nine numbers starting from 2, 3, 1, 512, 100473, or whatever. I wonder if it works for other progressions? Say, N⁵₅ = [5, 10, 15, ..., 45]? (It does, but proof is an exercise for the reader.) Or for negative integers? What would we have to do to the algorithm to make it work? What about magic squares of order 4, 5, 19? What about...?


Just look at all this number theory we got from a question where the suggested approach was to "fiddle randomly and hope you trip over the right answer."

I'm sure there's some sort of pedagogical approach which calls for the systematic frustration of children, and the comprehensive murder of any potential joy of mathematics, but for the life of me I can't think what it is.